26.9: The Merrifield Solid-Phase Technique
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- Steven Farmer, Dietmar Kennepohl, & William Reusch
- University of Illinois Springfield
Objectives
After completing this section, you should be able to describe, briefly, the Merrifield solid‑phase technique for the synthesis of polypeptides.
Key Terms
Make certain that you can define, and use in context, the key term below.
- solid‑phase method (solid‑phase synthesis)
Study Notes
The solid‑phase used in this method is a polymer support. You will not be examined on the details of the Merrifield solid‑phase method; however, you should be prepared to write a couple of paragraphs describing this important process.
For his work on the synthesis of peptides, Bruce Merrifield was awarded the 1984 Nobel Prize in chemistry.
The synthesis of a peptide of significant length (e.g. ten residues) by this approach requires many steps, and the product must be carefully purified after each step to prevent unwanted cross-reactions. To facilitate the tedious and time consuming purifications, and reduce the material losses that occur in handling, a clever modification of this strategy has been developed. This procedure, known as the Merrifield Synthesis after its inventor R. Bruce Merrifield , involves attaching the C-terminus of the peptide chain to a polymeric solid, usually having the form of very small beads. Separation and purification is simply accomplished by filtering and washing the beads with appropriate solvents. The reagents for the next peptide bond addition are then added, and the purification steps repeated. The entire process can be automated, and peptide synthesis machines based on the Merrifield approach are commercially available. A series of equations illustrating the Merrifield synthesis may be viewed below . The final step, in which the completed peptide is released from the polymer support, is a simple benzyl ester cleavage. This is not shown in the display.
| The Merrifield Peptide Synthesis | |
Two or more moderately sized peptides can be joined together by selective peptide bond formation, provided side-chain functions are protected and do not interfere. In this manner good sized peptides and small proteins may be synthesized in the laboratory. However, even if chemists assemble the primary structure of a natural protein in this or any other fashion, it may not immediately adopt its native secondary, tertiary and quaternary structure. Many factors, such as pH, temperature and inorganic ion concentration influence the conformational coiling of peptide chains. Indeed, scientists are still trying to understand how and why these higher structures are established in living organisms.


13.4: Chemical Shifts in ¹H NMR Spectroscopy
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Objectives
After completing this section, you should be able to
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state the approximate chemical shift (
δ
) for the following types of protons:
- aromatic.
- vinylic.
- those bonded to carbon atoms which are in turn bonded to a highly electronegative element.
- those bonded to carbons which are next to unsaturated centres.
- those bonded to carbons which are part of a saturated system.
- predict the approximate chemical shifts of each of the protons in an organic compound, given its structure and a table of chemical shift correlations.
Study Notes
You should not attempt to memorize the chemical shifts listed in the table of this section , although it is probable that you will need to refer to it quite frequently throughout the remainder of this course. To fulfill Objective 1, above, you should be familiar with the information presented in the figure of chemical shift ranges for organic compounds . If you have an approximate idea of the chemical shifts of some of the most common types of protons, you will find the interpretation of 1 H NMR spectra less arduous than it might otherwise be. Notice that we shall not try to understand why aromatic protons are deshielded or why alkynyl protons are not deshielded as much as vinylic protons. These phenomena can be explained, but the focus is on the interpretation of 1 H NMR spectra, not on the underlying theory.
1 H NMR Chemical Shifts
Chemical shifts in NMR (Nuclear Magnetic Resonance) spectroscopy refer to the phenomenon where the resonant frequency of a nucleus in a magnetic field is influenced by its chemical environment. This effect arises from the shielding or deshielding of the nucleus by the surrounding electron cloud. When a nucleus experiences different local electron densities due to nearby atoms or functional groups, its resonant frequency is altered relative to a reference standard, usually tetramethylsilane in organic solvents. This alteration is expressed in parts per million (ppm) and is known as the chemical shift. Tetramethylsilane [TMS;(CH 3 ) 4 Si] is generally used for standard to determine chemical shift of compounds: δ TMS =0 ppm. In other words, frequencies for chemicals are measured for a 1 H nucleus of a sample from the 1 H or resonance of TMS. It is important to understand trend of chemical shift in terms of NMR interpretation. The proton NMR chemical shift is affect by nearness to electronegative atoms (O, N, halogen.) and unsaturated groups (C=C,C=O, aromatic). Electronegative groups move to the down field (left; increase in ppm). Unsaturated groups shift to downfield (left) when affecting nucleus is in the plane of the unsaturation, but reverse shift takes place in the regions above and below this plane. 1 H chemical shift play a role in identifying many functional groups. Figure 1. indicates important example to figure out the functional groups.
Chemical shift values are in parts per million (ppm) relative to tetramethylsilane.
| Hydrogen type | Chemical shift (ppm) |
|---|---|
| RC H 3 | 0.9 - 1.0 |
| RC H 2 R | 1.2 - 1.7 |
| R 3 C H | 1.5 – 2.0 |
| 2.0 – 2.3 | |
| 1.5 – 1.8 | |
| RN H 2 | 1 - 3 |
| ArC H 3 | 2.2 – 2.4 |
| 2.3 – 3.0 | |
| ROC H 3 | 3.7 – 3.9 |
| 3.7 – 3.9 | |
| RO H | 1 - 5 |
| 3.7 – 6.5 | |
| 5 - 9 | |
| Ar H | 6.0 – 8.7 |
| 9.5 – 10.0 | |
| 10 - 13 |
Exercise \(\PageIndex{1}\)
The following have one H 1 NMR peak. In each case predict approximately where this peak would be in a spectra.
- Answer
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A. 5.20 δ; B. 1.50 δ; C. 6.40 δ; D. 1.00 δ
Exercise \(\PageIndex{2}\)
Identify the different equivalent protons in the following molecule and predict their expected chemical shift.
- Answer
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There are 6 different protons in this molecule
The shifts are (close) to the following: (a) 2 δ; (b) 6 δ; (c) 6.5 δ; (d) 7 δ; (e) 7.5 δ; (f) 7 δ


1.3: Habits for Success
- Define college success and examine strategies to achieve success in college
A Personal Definition for Success
How do you define college success? The definition really depends on you. You might think that “success” is earning an associate’s degree or attending classes in a four-year college. Maybe success is a bachelor’s or master’s degree or a Ph.D. Maybe success means receiving a certificate of completion or finishing skill-based training.
You might be thinking of other measures of college success, too—like grades. For instance, you might be unhappy with anything less than an A in a course, although maybe this depends on the difficulty of the subject. As long as you pass with a C, you might be perfectly content.
To help you start to define what success means to you, take this quick self-assessment about your college goals and beyond. How many of these items are important to you?
| I want to be able to . . . | YES | NO |
|---|---|---|
| Change my major during my college years | ||
| Have good relationships with my professors | ||
| Be eligible for financial aid | ||
| Be eligible for scholarships | ||
| Get good grades and awards | ||
| Get reductions on my car insurance | ||
| Prove to my employer that I can work hard | ||
| Keep my parents happy | ||
| Make social or professional connections | ||
| Get a better job after graduation |
Each of us has a different combination of answers to the questions above, but nearly every student in college answers yes to at least one of these questions. At the same time, we can’t be successful all the time at everything. We have to balance our energy and our focus, to get what we really want.
Deciding to go to college has an opportunity cost . An opportunity cost is based on the economic principle that there are limited resources available and choices must be made. Examples of resources would be things like time and money. If you are spending time doing something, you must give up doing something else you want to do. That is the opportunity cost of your choice. Going to college will have an opportunity cost in your life. An important question to ask at the beginning of your college venture is: what are you willing to trade-off for going to college?
Opportunity costs are tied to the idea of return on investment. Once you make an investment of your time and money in college, what investment are you hoping to get in return? How you define success in relation to your college experience impacts how you see the concept of return on investment. Some ways to gauge the return on your college investment include the following: job opportunities after college, financial benefits, social network/connections made while attending college, development of communication and other “soft skills,” and personal enrichment and/or happiness.
What kinds of trade-offs do you think succeeding in college will require of you? How do you think you might explain these trade-offs to someone who isn’t in college or never attended college?
Strategies to Achieve Success
Most students want to be successful in their courses, yet not all college students are. We’ve talked already about common misconceptions about not being “smart enough” for college. Some students suffer from impostor syndrome , or a feeling that they are impostors and don’t belong in college. Impostor syndrome is real and challenging.
Watch this clip about impostor syndrome to learn more about how it can affect college students and what to do about it.
Instead of worrying about whether you are smart enough or belong in college, you might focus on some concrete strategies you can employ to increase your success in college. We suggest you embrace and master the following eight strategies:
- Learn how to take effective notes in class.
- Review the text and your reading notes prior to class.
- Participate in class discussions.
- Go to office hours and ask your instructor questions.
- Give yourself enough time to research, write, and edit your essays in manageable stages.
- Take advantage of online or on-campus academic support resources.
- Spend sufficient time studying.
- Join a study group or find a study buddy.
If you find yourself worrying about succeeding in college, ask yourself instead if you can implement or improve your use of some of these strategies.
Can you make more time for learning? One approach is to create a regular study schedule and make sure you allow yourself ample time. Only break away from your committed schedule if an extreme situation prevents you from sticking to it.
Most college success experts agree that students should study two hours outside of class for every hour in class.
Another strategy to consider implementing is group study. For example, rather than relying just on your own knowledge, notes, and skills, try studying with other students in your classes. Studying in a group or with a buddy gives you a chance to ask questions and talk about concepts.
You can also add a tutor to your personal study or to your study group. Tutoring is generally free in college, and the strategies and knowledge you gain will be invaluable. Usually, tutors have taken the class you are currently enrolled in, and they are trained to get the best out of you.
Overall, students struggle in college not because of natural intellect or smarts, but because of poor time management, disorganization, and lack of quality study time. The good news is that there are ways to combat this, specifically by doing things like creating a regular study schedule, studying in groups, and taking advantage of your school’s academic resources, such as tutoring centers, instructor office hours, and online help.
This Ted talk video by Alain de Botton stresses the external factors that enable “success” or “failure” and encourages us to try to resist society’s definitions of success and strive to author our own ambitions. Click here to download a transcript for this video.
impostor syndrome : impostor syndrome is the consistent fear that one does not belong or that one is not good enough; someone suffering from impostor syndrome may worry about being “exposed” as a fraud or fake
opportunity cost : the “cost” of choosing one alternative versus another; if I go running instead of hanging out with my friends, I incur the opportunity cost of not socializing
office hours: office hours are a specific time faculty set aside to meet individually with students who want extra help, have questions, or seek mentoring; some faculty ask students to make an appointment while others have drop-in office hours
Contributors and Attributions
- Impostor syndrome. Authored by : J. Emilio Flores. Located at : https://vimeo.com/208883682 . License : Public Domain: No Known Copyright
- Revision, Adaptation, and Original Content. Provided by : Lumen Learning. License : CC BY: Attribution
- College Success. Authored by : Linda Bruce. Provided by : Lumen Learning. Located at : courses.lumenlearning.com/lumencollegesuccess. License : CC BY: Attribution
- Image of woman smiling. Authored by : Felip1. Located at : https://flic.kr/p/riat9A . License : CC BY-NC: Attribution-NonCommercial
- License : CC BY: Attribution
- Image of four students. Authored by : UBC Library Communications. Located at : flic.kr/p/57H7r4. License : CC BY-NC-ND: Attribution-NonCommercial-NoDerivatives
- License : CC BY: Attribution
- Foundations of College Success: Words of Wisdom. Authored by : Thomas C. Priester, editor. Provided by : Open SUNY Textbooks. Located at : textbooks.opensuny.org/foundations-of-academic-success/. License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike
- Alain de Botton: A kinder, gentler philosophy of success. Authored by : TED. Located at : https://youtu.be/MtSE4rglxbY . License : All Rights Reserved . License Terms : Standard YouTube License


5.1: Module 22 Assemblies
22
Module 22 Assemblies
Wally Baumback
Learning Outcomes
When you have completed this module, you will be able to:
- Describe an assembly and explain the difference between a top-down and a bottom-up assembly.
- Describe and apply the SLICE GRAPHICS command.
- Describe the PLACE COMPONENT and PLACE CONSTRAINT commands and apply them to assemble a series of parts to create an assembled model.
An Assembly File
An assembly file contains the information required to assemble two or more part files to create an assembled model. See Figure 22-1. As the model is assembled, assembly constraints must be assigned so that each part knows how it aligns or fits with the other parts in the assembled model. A part (.ipt) file that has been placed in the assembly can be edited while in the assembly file and the modifications will be saved back to the original part file. On the other hand, if the original part file is modified after the assembly file has being created, the modifications will automatically display in the assembly. An assembly file does not actually contain any of the part files that are placed in the file, it simply contains a reference to them.
A reference is a link back to the part files. The part files that have been placed in an assembly file must be available to Inventor to display them when an assembly file is opened. The current project file keeps track of those links and will automatically keep track of the location of all the part files. If an assembly file is sent to a client or an associate, the part files that were placed in the assembly file must also be included.
Creating an Assembly File
To create an assembly file, use the NEW command and select an assembly template rather than a part template. An assembly file has the extension .iam. IAM is an acronym for Inventor Assembly.
MUST KNOW: An assembly file contains the information required to assemble two or more part files to create the assembled model. An assembly file has the file extension .iam. IAM is an acronym for Inventor Assembly.Bottom-up vs Top-down
Assembly files can be created using either the bottom-up or the top-down method. A bottom-up assembly is an assembly created from a series of part files that were previously created and saved in their own .ipt file. A top-down assembly is an assembly file where all parts of the assembly are created on the fly. In other word, they are created in the active assembly file, one at a time. They are aligned and constrained in their correct position in relationship to the other parts in the assembled model. Inventor will save each part in its own file with the extension .ipt. In the Inventor book, only the bottom-up method is taught.
The Browser Bar in an Assembly
When an assembly is the active file, the Browser bar will display all of the parts that have been placed in the assembly. If a part in the browser is expanded it will display its children which includes its alignment and constraints to the other parts of the assembled model. See Figure 22-2.
If the same part is placed more then once into the assembly file, the Browser bar will number them accordingly. For example, this may happen if an assembly required two identical bolts. Only one part file is created and named Bolt.ipt. It is then placed into the assembly file twice. The Browser bar would number the bolts parts as follows:
On the first occurrence: Bolt:1
On the second occurrence: Bolt:2
Inventor Command: SLICE GRAPHICS
The SLICE GRAPHICS command is used to slice away the model on the active sketching plane temporarily removing all of the material in front of the sketching plane. The sketching plane must be active before using this command.Shortcut: F7
WORK ALONG: Creating the Parts for an Assembly
Step 1
Check the default project and if necessary, set it to Inventor Course.
Step 2
Use the following instructions to complete all parts in this workalong. Create the following parts and ensure that you do the following:
A Each part must be saved in its own .ipt file.
B Project the Center Point onto the Base sketch plane and note the location of X0Y0Z0.
C
Draw the necessary sketches and extrude or revolve them to produce the solid model shown. Apply all of the necessary geometrical and dimensional constraints to fully constrain each sketch.
D
Apply the colour and material shown.
Step 3
Construct Part A. (Figure Step 3A, 3B, and 3C)
Part: Base
Part Name: Inventor Workalong 22-1A
Template: English-Modules Part (in).ipt
Color: Stainless – Brushed
Material: Stainless Steel
Step 4
Construct Part B. (Figure Step 4A and 4B)
Note: See Steps 3 to 14 if you require help creating this model.
Part: Post
Part Name: Inventor Workalong 22-1B
Template: English-Modules Part (in).ipt
Color: Chrome – Black Polished
Material: Steel
Step 5
Draw the Base model of part 22-1B by extruding circles. Your Base model should appear as shown in the figure. (Figure Step 5)
Step 6
Start a new sketch on the top of the model and draw a construction line from the centre to the edge of the top plane along the Y axis as shown in the figure. Hint: Ensure that you draw the line along the Y axis by snapping to the centre of the circle and to the grid on the Y axis. The length of the line is not important. (Figure Step 6)
Step 7
Insert a Work Plane in the centre of the model using the Perpendicular to a Line method. Enter the WORK PLANE command and move the cursor to the end of the construction line. The Work Plane icon will display as shown in the figure. Click the end of the line when the icon displays. (Figure Step 7)
USER TIP: Before drawing each part, the orientation of the part when it is placed in the future assembly, should be considered. Doing that, will save a lot of time when creating the assembly model since it can be placed without any manipulation.Step 8
Move the cursor onto the Y axis and it will display as shown in the figure. When the Y axis is displayed, click it and the Work Plane will display as shown. (Figure Step 8A and 8B)
Step 9
Enlarge the Work Plane to extend it past the edges of the model. One way to do this is to use the LOOK AT/VIEW FACE command and change the view so that it is looking perpendicular to the plane as shown in the figure. (Figure Step 9)
Step 10
Start a new sketch on the Work Plane. (Figure 10)
Step 11
Enter the SLICE GRAPHIC command (F7) and the model will display from the sketching plane back. Disable the grid display. It is easier to draw this sketch without the grid. (Figure Step11)
USER TIP: When enlarging the work plane, the cursor must display as the Stretch Move Stretch Cursor (two arrows) as shown. If you Cursor Cursor have trouble changing the cursor from the Move Cursor, zoom in closer and move the cursor onto one of the corners of the plane.Step 12
Using the PROJECT GEOMETRY command, project the Z axis and the X axis onto the sketching plane. (Figure Step 12A and 12B)
Step 13
Using the OFFSET command, construct offsets to both the Z and X axis to start the construction of the slot. Trim the lines in the sketch and insert the dimensions to fully constrain the sketch. Ensure that you dimension from both sides of the Z axis as shown in the figure. (Figure Step 13A and 13B)
MUST KNOW: An assembly file does do not contain any of the part files that are placed in the assembled model. It simply contains a reference to the part files. A reference is a link back to the part files. The part files that have been placed in an assembly file must be available to Inventor to display them when the assembly file is opened.The project file keeps track of those links and will automatically keep track of the location of all the files. If an assembly file (.iam) is sent to a client or an associate, the part files (.ipt) that were placed in the assembly file must also be included otherwise they will not display when the file is opened.
Step 14
Extrude the sketch. Set it to extrude in both directions and to cut. Set the distance to All. Complete the part by adding the hole, the threads, the fillets, the colour, and the material as specified in Figure 14A, 14B, and 14C.
Step 15
Create Part C. (Figure Step 15A, 15B, 15C, and 15D):
Part: Wedge Ring
Part Name:
Inventor Workalong 22-1C
Template:
English-Modules Part (in).ipt
Color:
Titanium – Polished
Material:
Titanium
Step 16
Construct Part D. (Figure Step 16A and 16B)
Part:
Wedge
Part Name:
Inventor Workalong 22-1D
Template:
English-Modules Part (in).ipt
Color:
Chrome – Polished
Material:
Steel
Step 17
Change the face colour of the top face of the part to: Metal Steel (Knurled) as shown in Figure 16A.
Step 18
Construct Part F as follows: (Figure Step 18A and 18B)
Part:
Handle
Part Name:
Inventor Workalong 22-1E
Template:
English-Modules Part (in).ipt
Color:
Metal-AL-6061 – Machined
Material:
Steel
Step 19
Construct Part E as follows: (Figure Step 19A, 19B, and 19C)
Part:
Screw
Part Name:
Inventor Workalong 22-1F
Template:
English-Modules Part (in).ipt
Color:
Metal-AL-6061 – Machined
Material:
Steel
Step 20
Change the face colour of the cylindrical top of the part to: Metal Steel (Knurled) as shown in the solid model figure.
Inventor Command: PLACE COMPONENT
The PLACE COMPONENT command is used to insert a part or a component into an assembly file.
Shortcut: P
Inventor Command: PLACE CONSTRAINT
The PLACE CONSTRAINT command is used to apply constraints of one part to another in the assembled model.
Shortcut: C
Grounded Parts
By default, the first part that is placed into an assembly file will be grounded. A grounded part is a part that has all of it degrees of freedom removed and is fully constrained in that file. It is important to ground at least one part of every assembly. If no parts are grounded, the assembly can be moved around in model space. Once one part is grounded, the other parts can be constrained to it making their movement relative to the grounded part. If required, more then one part can be grounded.
In the Browser bar, a grounded part will display a Push Pin icon as shown in Figure 22-3 and 22-4. To enable or disable a part being grounded, right-click the part in the Browser bar. In the Right-click menu, select Grounded. In Figure 22-5, the selected part is currently grounded, enabled and visible.
Part Visibility
The visibility of parts in an assembly can be enabled or disabled. When the part’s visibility is disabled in the assembly, the part’s icon in the Browser bar will display greyed out.
Enabled Parts
Parts can be enabled or disabled in the assembly file. If a part is enabled, it displays and can be selected in the assembly. If it is disabled, only an outline displays and It cannot be selected. Sometimes it is easier to disable some parts to make it easier to place additional parts.
Assembly Constrains
There are many different constraints used when creating an assembly. In this module, only the mate constraint will be taught. A mate constraint constrains two assembled parts to one another by mating their centerlines and/or by mating a face on one part to a face on the other part. Mating is the most common way to assemble two parts together. It often takes more then one constraint to assemble two parts together. Both parts must have a symmetrical feature to mate them using the centerline method.
MUST KNOW: By default, the first part that is placed into an assembly file will be grounded. A grounded part is a part that has all of it degrees of freedom removed and is fully constrained in that file. It is important to ground at least one part in every assembly since if no parts are grounded, the assembly can be moved around model space. Once one part is grounded, the other parts can be constrained to it making their movement relative to the grounded part. USER TIP: The Browser bar contains a great deal of information about the active assembly file. The grounded part has a Push Pin icon. If a part is disabled, it will display a green part icon and if its visibility is disabled, the part icon will be grey as you can see in the figure.WORKALONG: Creating an Assembly
Step 1
Check the default project and if necessary, set it to Inventor Course.
Step 2
Using the NEW command, enable the English tab and select the temple file: Modules Assembly (in).iam . Save and name the assembly: Inventor workalong 22-1 . (Figure Step 2)
Step 3
Enter the PLACE COMPONENT command. In the Place Component dialogue box insert part files: Inventor Workalong 22-1A.ipt and Inventor Workalong 22-1B.ipt into the assembly file. (Figure Step 3A and 3B)
USER TIP: By default, the PLACE COMPONENT command places multiple parts. To insert the part more than once into the assembly file, click once for each part. Ensure that you move the cursor so you are not inserting one part on top of another one. When you have inserted the required part(s), press Esc to exit the command.Step 4
Disable the grounding of part 22-1A and enable the grounding of part 22-1B. (Figure Step 4A, 4B, and 4C)
Step 5
Enter the PLACE CONSTRAINT command. In the Place Constraint dialogue box set Type to Mate, Selections to 1, Offset to 0.000 and Solution to Mate. (Figure Step 5)
Step 6
Move the cursor onto the part 22-1A and move it until the Centerline constraint symbol appears. When it displays as shown in the figure, select it. Move the cursor onto the part 22-1B and do the same. Part 22-1A will move onto Part 22-1B. Click Apply. (Figure Step 6A, 6B, and 6C)
Step 7
Move the cursor onto the top of the base on part 22-1B as shown in the figure. The mate arrow will display pointing up. Press F4 and rotate the model. Select the mate constraint for the bottom of the part 22-1A. (Figure Step 7A, 7B, and 7C)
Step 8
Press F6 to change the Home view. (Figure Step 8)
Step 9
Enter the PLACE COMPONENT (P) command and insert part 22-1C, 22-1D, 22-1E, and 22-1F. Ensure that you place each part only once. (Figure Step 9)
Step 10
Using what you just learned, assemble and constrain part 22-1C. (Figure Step 10)
Step 11
Using the PLACE CONSTRAINT command, constrain the arc in part 22-1D to the arc in part 22-1C. (Figure Step 11)
Step 12
In the Offset box, set the offset to 0.015. Enable the Pick Part First box. See the Author’s Comments below. (Figure Step 12)
Step 13
To constrain the wedge in the slot, select one side of the slot and one side of the wedge as the mate surfaces. (Figure Step 13A, 13B, and 13C)
Step 14
Using what you just learned, constrain the last two parts to complete the assembly. (Figure Step 14)
Step 15
Save and close the file.
Key Principles
Key Principles in Module 22
- An assembly file contains the information required to assemble two or more part files to create the assembled model. An assembly file has the file extension .iam.
- An assembly file does do not contain any of the part files that are placed in the assembled model. It simply contains a reference to the part files. If an assembly file (.iam) is sent to a client or an associate, the part files (.ipt) that were placed in the assembly file must also be included otherwise they will not display when the assembly file is opened.
- By default, the first part that is placed into an assembly file will be grounded. A grounded part is a part that has all of it degrees of freedom removed and is fully constrained in that file.
- There are many different constraints used when creating an assembly. A mate constraint constrains two assembled parts to one another by mating their centerlines and/or by mating a plane on one part to a plane on the other part.
Lab Exercise 22-1
Time allowed: 180 minutes.
| Part Name | Project | Units | Template | Color | Material |
| Inventor Lab 22-1 | Inventor Course | Millimeters | See Below | See Below | See Below |
Step 1
Create the following parts.
Step 2
Each part must have its own file. In each part, ensure that you do the following:
A
Select your own location of X0Y0Z0.
B
Draw the necessary sketches and extrude or revolve them to produce the solid model shown. Apply all of the necessary geometrical and dimensional constraints to fully constrain all sketches. (Figure Step 2A and 2B)
C Apply the colour and material shown.
Part: Tire
Part Name:
Inventor Lab 22-1A
Template:
Metric-Modules Part (mm).ipt
Color:
Rubber – Green – Version 1.1
Material:
Rubber
Part:
Frame
Part Name:
Inventor Lab 22-1B
Template:
Metric-Modules Part (mm).ipt
Color:
Aluminum – Polished
Material:
Steel (Figure Step 2C and 2D)
Part:
Bushing
Part Name:
Inventor Lab 22-1C
Template:
Metric-Modules Part (mm).ipt
Color:
Brass – Satin
Material:
Brass (Figure Step 2E and 2F)
Part:
Pin
Part Name:
Inventor Lab 22-1D
Template:
Metric-Modules Part (mm).ipt
Color:
Semi – Polished
Material:
Steel (Figure Step 2G, 2H and 2J)
Part:
10 mm Nut
Part Name:
Inventor Lab 22-1E
Template:
Metric-Modules Part (mm).ipt
Color:
Semi – Polished
Material:
Steel (Figure Step 2K and 2L)
Part:
Bolt
Part Name:
Inventor Lab 22-1F
Template:
Metric-Modules Part (mm).ipt
Color:
Semi – Polished
Material:
Steel (Figure Step 2M and 2N)
Part:
12 mm Nut
Part Name:
Inventor Lab 22-1G
Template:
Metric-Modules Part (mm).ipt
Color:
Semi-Polished
Material:
Steel (Figure Step 2P and 2Q)
Part:
Washer
Part Name:
Inventor Lab 22-1H
Template:
Metric-Modules Part (mm).ipt
Color:
Semi-Polished
Material:
Steel (Figure Step 2R and 2S)
Step 3
Open a new assembly file.
Assembly:
Caster
File Name:
Inventor Lab 22-1.iam
Template:
Metric – Modules Assembly (mm).iam
Step 4
Set the Tire as the grounded part.
Step 5
Start with assembling the Bushing into the Tire first. (Figure Step 5)
Step 6
Assemble the remaining parts to complete the assemble file. (Figure Step 6A and 6B)


19.8: Technology and New Infectious Agents
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Learning Objectives
- Give examples demonstrating the positive and negative impacts technology has had on new infectious agents
Technology: The Good
The use of advanced technology and molecular methods for detection, identification, and characterization of infectious agents is gaining importance in clinical microbiology laboratories. Emerging and re-emerging pathogens pose several challenges to diagnosis, treatment, and public health surveillance. Identification of an emerging pathogen by conventional methods is difficult and time-consuming due to the ‘novel’ nature of the agent. Identification requires a large array of techniques including cell cultures, inoculation of animals, cultivation using artificial media, histopathological evaluation of tissues (if available), and serological techniques using surrogate antigens.
Looking back at past epidemics or outbreaks caused by previously unknown infectious agents, we realize that identification and characterization of a new infectious agent can take years, decades, or even centuries. Such time frames have been decreased to weeks or months by the use of powerful molecular techniques, as seen with the identification of severe acute respiratory syndrome coronavirus (SARS-CoV) within weeks of the first cases reported, the discovery of a new hantavirus in North America in 1993, and the detection of bacteria as etiological pathogens of human infections such as Ehrlichia chaffeensis and Anaplasma phagocytophilum in human monocytotropic ehrlichiosis and human granulocytotropic anaplasmosis, respectively.
Molecular techniques offer several advantages over conventional methods, including high sensitivity and specificity, speed, ease of standardization, and automation. Other advantages include identification of novel, non-cultivable or very slowly growing organisms, strain typing in epidemiological studies, antimicrobial susceptibility determination, and monitoring treatment by measuring bacterial or viral loads.
Technology: The Bad
The effects of new technology on the environment are related to the emergence of many infectious diseases. For example, Lyme disease, hantavirus pulmonary syndrome (HPS), and Lassa fever all emerged when humans began encountering the insect vector (for Lyme disease) or rodent host (for HPS and Lassa fever) of the causative agents in greater numbers than ever before. Factors related to the emergence of infectious diseases such as Legionnaire disease and hemolytic uremic syndrome include changing technologies: air conditioning systems and mass food production, respectively.
Technology: The Ugly
Several human activities have led to the emergence and spread of new diseases:
ENCROACHMENT ON WILDLIFE HABITATS
The construction of new villages and housing developments in rural areas forces animals to live in dense populations, creating opportunities for microbes to mutate and emerge.
CHANGES IN AGRICULTURE
The introduction of new crops attracts new crop pests and the microbes they carry to farming communities, exposing people to unfamiliar diseases.
DESTRUCTION OF RAIN FORESTS
As countries make use of their rain forests by building roads and clearing areas for settlement or commercial ventures, people encounter insects and other animals harboring previously unknown microorganisms.
UNCONTROLLED URBANIZATION
The rapid growth of cities in many developing countries tends to concentrate large numbers of people into crowded areas with poor sanitation. These conditions foster transmission of contagious diseases.
MODERN TRANSPORT
Ships and other cargo carriers often harbor unintended ‘passengers’ that can spread diseases to faraway destinations. With international air travel, people infected with a disease can carry it to distant lands, or home to their families, before their first symptoms appear.
Key Points
- Advanced technology enables rapid identification of pathogens causing disease outbreaks and helps accelerate treatment strategies.
- The effect of new technology on the environment is related to the emergence of many new infectious diseases.
- Infectious diseases are sometimes called ” contagious ” when they are easily transmitted by contact with an ill person or their secretions (e.g., influenza).
Key Terms
- infectious : Infectious diseases, also known as transmissible diseases or communicable diseases, comprise clinically evident illness (i.e., characteristic medical signs and/or symptoms of disease) resulting from the infection, presence, and growth of pathogenic biological agents in an individual host organism.
- pathogens : A pathogen or infectious agent (colloquially known as a germ) is a microorganism (in the widest sense, such as a virus, bacterium, prion, or fungus) that causes disease in its host. The host may be an animal (including humans), a plant, or even another microorganism.


6.S: Direct-Current Circuits (Summary)
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Key Terms
| ammeter | instrument that measures current |
| electromotive force (emf) | energy produced per unit charge, drawn from a source that produces an electrical current |
| equivalent resistance | resistance of a combination of resistors; it can be thought of as the resistance of a single resistor that can replace a combination of resistors in a series and/or parallel circuit |
| internal resistance | amount of resistance to the flow of current within the voltage source |
| junction rule | sum of all currents entering a junction must equal the sum of all currents leaving the junction |
| Kirchhoff’s rules | set of two rules governing current and changes in potential in an electric circuit |
| loop rule | algebraic sum of changes in potential around any closed circuit path (loop) must be zero |
| potential difference | difference in electric potential between two points in an electric circuit, measured in volts |
| potential drop | loss of electric potential energy as a current travels across a resistor, wire, or other component |
| RC circuit | circuit that contains both a resistor and a capacitor |
| shock hazard | hazard in which an electric current passes through a person |
| terminal voltage | potential difference measured across the terminals of a source when there is no load attached |
| thermal hazard | hazard in which an excessive electric current causes undesired thermal effects |
| three-wire system | wiring system used at present for safety reasons, with live, neutral, and ground wires |
| voltmeter | instrument that measures voltage |
Key Equations
| Terminal voltage of a single voltage source | \(V_{terminal}=ε−Ir_{eq}\) |
| Equivalent resistance of a series circuit | \(R_{eq}=R_1+R_2+R_3+⋯+R_{N−1}+R_N=\sum_{i=1}^NR_i\) |
| Equivalent resistance of a parallel circuit | \(R_{eq}=(\frac{1}{R_1}+\frac{1}{R_2}+⋯+\frac{1}{R_N})^{−1}=(\sum_{i=1}^N\frac{1}{R_i})^{−1}\) |
| Junction rule | \(\sum I_{in}=\sum I_{out}\) |
| Loop rule | \(\sum V=0\) |
| Terminal voltage of N voltage sources in series | \(V_{terminal}=\sum_{i=1}^Nε_i−I\sum_{i=1}^Nr_i=\sum_{i=1}^Nε_i−Ir_{eq}\) |
| Terminal voltage of N voltage sources in parallel | \(V_{terminal}=ε−I\sum_{i=1}^N(\frac{1}{r_i})^{−1}=ε−Ir_{eq}\) |
| Charge on a charging capacitor | \(q(t)=Cε(1−e^{−\frac{t}{RC}})=Q(1−e^{−\frac{t}{τ}})\) |
| Time constant | \(τ=RC\) |
| Current during charging of a capacitor | \(I=\frac{ε}{R}e^{−\frac{t}{RC}}=I_oe^{−\frac{t}{RC}}\) |
| Charge on a discharging capacitor | \(q(t)=Qe^{−\frac{t}{τ}}\) |
| Current during discharging of a capacitor | \(I(t)=−\frac{Q}{RC}e^{−\frac{t}{τ}}\) |
Summary
10.2 Electromotive Force
- All voltage sources have two fundamental parts: a source of electrical energy that has a characteristic electromotive force (emf), and an internal resistance r. The emf is the work done per charge to keep the potential difference of a source constant. The emf is equal to the potential difference across the terminals when no current is flowing. The internal resistance r of a voltage source affects the output voltage when a current flows.
- The voltage output of a device is called its terminal voltage \(V_{terminal}\) and is given by \(V_{terminal}=ε−Ir\), where I is the electric current and is positive when flowing away from the positive terminal of the voltage source and r is the internal resistance.
10.3 Resistors in Series and Parallel
- The equivalent resistance of an electrical circuit with resistors wired in a series is the sum of the individual resistances:
\(R_s=R_1+R_2+R_3+⋯=\sum_{i=1}^NR_i\).
- Each resistor in a series circuit has the same amount of current flowing through it.
- The potential drop, or power dissipation, across each individual resistor in a series is different, and their combined total is the power source input.
- The equivalent resistance of an electrical circuit with resistors wired in parallel is less than the lowest resistance of any of the components and can be determined using the formula
\(R_{eq}=(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+⋯)^{−1}=(\sum_{i=1}^N\frac{1}{R_i})^{−1}\).
- Each resistor in a parallel circuit has the same full voltage of the source applied to it.
- The current flowing through each resistor in a parallel circuit is different, depending on the resistance.
- If a more complex connection of resistors is a combination of series and parallel, it can be reduced to a single equivalent resistance by identifying its various parts as series or parallel, reducing each to its equivalent, and continuing until a single resistance is eventually reached.
10.4 Kirchhoff's Rules
- Kirchhoff’s rules can be used to analyze any circuit, simple or complex. The simpler series and parallel connection rules are special cases of Kirchhoff’s rules.
- Kirchhoff’s first rule, also known as the junction rule, applies to the charge to a junction. Current is the flow of charge; thus, whatever charge flows into the junction must flow out.
- Kirchhoff’s second rule, also known as the loop rule, states that the voltage drop around a loop is zero.
- When calculating potential and current using Kirchhoff’s rules, a set of conventions must be followed for determining the correct signs of various terms.
- When multiple voltage sources are in series, their internal resistances add together and their emfs add together to get the total values.
- When multiple voltage sources are in parallel, their internal resistances combine to an equivalent resistance that is less than the individual resistance and provides a higher current than a single cell.
- Solar cells can be wired in series or parallel to provide increased voltage or current, respectively.
10.5 Electrical Measuring Instruments
- Voltmeters measure voltage, and ammeters measure current. Analog meters are based on the combination of a resistor and a galvanometer, a device that gives an analog reading of current or voltage. Digital meters are based on analog-to-digital converters and provide a discrete or digital measurement of the current or voltage.
- A voltmeter is placed in parallel with the voltage source to receive full voltage and must have a large resistance to limit its effect on the circuit.
- An ammeter is placed in series to get the full current flowing through a branch and must have a small resistance to limit its effect on the circuit.
- Standard voltmeters and ammeters alter the circuit they are connected to and are thus limited in accuracy.
- Ohmmeters are used to measure resistance. The component in which the resistance is to be measured should be isolated (removed) from the circuit.
10.6 RC Circuits
- An RC circuit is one that has both a resistor and a capacitor.
- The time constant \(τ\) for an RC circuit is \(τ=RC\).
- When an initially uncharged (\(q=0\) at \(t=0\)) capacitor in series with a resistor is charged by a dc voltage source, the capacitor asymptotically approaches the maximum charge.
- As the charge on the capacitor increases, the current exponentially decreases from the initial current: \(I_0=ε/R\).
- If a capacitor with an initial charge Q is discharged through a resistor starting at \(t=0\), then its charge decreases exponentially. The current flows in the opposite direction, compared to when it charges, and the magnitude of the charge decreases with time.
10.7 Household Wiring and Electrical Safety
- The two types of electric hazards are thermal (excessive power) and shock (current through a person). Electrical safety systems and devices are employed to prevent thermal and shock hazards.
- Shock severity is determined by current, path, duration, and ac frequency.
- Circuit breakers and fuses interrupt excessive currents to prevent thermal hazards.
- The three-wire system guards against thermal and shock hazards, utilizing live/hot, neutral, and ground wires, and grounding the neutral wire and case of the appliance.
- A ground fault circuit interrupter (GFCI) prevents shock by detecting the loss of current to unintentional paths.


4.5: Nervous Tissue Mediates Perception and Response
Learning Objectives
- Identify the classes of cells that make up nervous tissue
- Discuss how nervous tissue mediates perception and response
Nervous tissue is characterized as being excitable and capable of sending and receiving electrochemical signals that provide the body with information. Two main classes of cells make up nervous tissue: the neuron and neuroglia (Figure \(\PageIndex{1}\)). Neurons propagate information via electrochemical impulses, called action potentials, which are biochemically linked to the release of chemical signals. Neuroglia play an essential role in supporting neurons and modulating their information propagation.
Exercise \(\PageIndex{1}\)
Follow this link to learn more about nervous tissue. What are the main parts of a nerve cell?
www.nobelprize.org/education...ng.html#/plot2
Neurons display distinctive morphology, well suited to their role as conducting cells, with three main parts. The cell body includes most of the cytoplasm, the organelles, and the nucleus. Dendrites branch off the cell body and appear as thin extensions. A long “tail,” the axon, extends from the neuron body and can be wrapped in an insulating layer known as myelin , which is formed by accessory cells. The synapse is the gap between nerve cells, or between a nerve cell and its target, for example, a muscle or a gland, across which the impulse is transmitted by chemical compounds known as neurotransmitters. Neurons categorized as multipolar neurons have several dendrites and a single prominent axon. Bipolar neurons possess a single dendrite and axon with the cell body, while unipolar neurons have only a single process extending out from the cell body, which divides into a functional dendrite and into a functional axon. When a neuron is sufficiently stimulated, it generates an action potential that propagates down the axon towards the synapse. If enough neurotransmitters are released at the synapse to stimulate the next neuron or target, a response is generated.
The second class of neural cells comprises the neuroglia or glial cells, which have been characterized as having a simple support role. The word “glia” comes from the Greek word for glue. Recent research is shedding light on the more complex role of neuroglia in the function of the brain and nervous system. Astrocyte cells , named for their distinctive star shape, are abundant in the central nervous system. The astrocytes have many functions, including regulation of ion concentration in the intercellular space, uptake and/or breakdown of some neurotransmitters, and formation of the blood-brain barrier, the membrane that separates the circulatory system from the brain. Microglia protect the nervous system against infection but are not nervous tissue because they are related to macrophages. Oligodendrocyte cells produce myelin in the central nervous system (brain and spinal cord) while the Schwann cell produces myelin in the peripheral nervous system (Figure \(\PageIndex{2}\)).
Figure \(\PageIndex{2}\): Nervous Tissue.  The cells of nervous tissue are specialized to transmit and receive impulses. LM × 872. (Micrograph provided by the Regents of University of Michigan Medical School © 2012)
Chapter Review
The most prominent cell of the nervous tissue, the neuron, is characterized mainly by its ability to receive stimuli and respond by generating an electrical signal, known as an action potential, which can travel rapidly over great distances in the body. A typical neuron displays a distinctive morphology: a large cell body branches out into short extensions called dendrites, which receive chemical signals from other neurons, and a long tail called an axon, which relays signals away from the cell to other neurons, muscles, or glands. Many axons are wrapped by a myelin sheath, a lipid derivative that acts as an insulator and speeds up the transmission of the action potential. Other cells in the nervous tissue, the neuroglia, include the astrocytes, microglia, oligodendrocytes, and Schwann cells.
Interactive Link Questions
Follow this link to learn more about nervous tissue. What are the main parts of a nerve cell?
Answer: Dendrites, cell body, and the axon.
Review Questions
Q. The cells responsible for the transmission of the nerve impulse are ________.
A. neurons
B. oligodendrocytes
C. astrocytes
D. microglia
Answer: A
Q. The nerve impulse travels down a(n) ________, away from the cell body.
A. dendrite
B. axon
C. microglia
D. collagen fiber
Answer: B
Q. Which of the following central nervous system cells regulate ions, regulate the uptake and/or breakdown of some neurotransmitters, and contribute to the formation of the blood-brain barrier?
A. microglia
B. neuroglia
C. oligodendrocytes
D. astrocytes
Answer: D
Critical Thinking Questions
Q. Which morphological adaptations of neurons make them suitable for the transmission of nerve impulse?
A. Neurons are well suited for the transmission of nerve impulses because short extensions, dendrites, receive impulses from other neurons, while a long tail extension, an axon, carries electrical impulses away from the cell to other neurons.
Q. What are the functions of astrocytes?
A. Astrocytes regulate ions and uptake and/or breakdown of some neurotransmitters and contribute to the formation of the blood-brain-barrier.
References
Stern, P. Focus issue: getting excited about glia. Science [Internet]. 2010 [cited 2012 Dec 4]; 3(147):330-773. Available from:
http://stke.sciencemag.org/cgi/conte...ans;3/147/eg11
Ming GL, Song H. Adult neurogenesis in the mammalian central nervous system. Annu. Rev. Neurosci. 2005 [cited 2012 Dec 4]; 28:223–250.
Glossary
- astrocyte
- star-shaped cell in the central nervous system that regulates ions and uptake and/or breakdown of some neurotransmitters and contributes to the formation of the blood-brain barrier
- myelin
- layer of lipid inside some neuroglial cells that wraps around the axons of some neurons
- neuroglia
- supportive neural cells
- neuron
- excitable neural cell that transfer nerve impulses
- oligodendrocyte
- neuroglial cell that produces myelin in the brain
- Schwann cell
- neuroglial cell that produces myelin in the peripheral nervous system


5.1: Introduction to Sampling Distributions
- Define and construct probability distributions of sample statistics with simple random sampling
- Define and construct sampling distributions of sample statistics
- Define and give examples of unbiased estimators
- Explore the impact sample size has on the sampling distributions of a given population
- Represent sampling distributions as continuous random variables
Review and Preview
In the previous chapter, we began our study of random variables by briefly connecting them with inferential and sample statistics. This chapter studies sample statistics as random variables, paying close attention to probability distributions. Recall for each random variable, an underlying random experiment will be conducted. A value is assigned, measured, or computed for each possible outcome of this random experiment. These are the values the random variable is said to take on, and the probability of the values occurring is determined, forming a probability distribution. If the random variable is discrete, each value has a specific probability. If the random variable is continuous, a range of values has a probability based on the area under the probability density function. We can compute the expected value (mean), variance, and standard deviation to help describe the random variable.
Suppose we wanted to know the average height of an American. It is impossible to measure the height of everyone in the country, so we decided to measure the heights of 10,000 randomly selected Americans and found that the average height of these 10,000 people is \(68.4\) inches. We cannot conclude that the average height of all Americans is \(68.4\) inches. After all, we may have selected the 10,000 tallest people in the country, or perhaps 8,000 were well below the average height. Even if our sampling method is unbiased, there is still the possibility that we obtained a sample mean far away from the population mean by pure chance. Even if the sample mean is close to the population mean, it likely is not the same. Do we need to measure more heights? Can 10,000 people accurately represent the whole country? To understand the average height of an American, we need to determine the probability that our sample mean is not accurate. What is the probability that our sample mean is off by more than \(1\) inch? What is the probability that it is off by less than \(0.1\) inches? To answer these questions, we need to think of sampling as a random variable.
Now that we have looked at the basics of random variables and have an example in mind, we study sample statistics as random variables in depth. We are interested in learning the characteristics of a population (parameters). Studying the entire population may be impossible, too expensive, or time-consuming, so we study a sample and compute a statistic to estimate the parameter. Ideally, the sample is representative of the population; it does not misrepresent the population, and the statistic is close to the parameter. We cannot guarantee such a sample, but by choosing our sample randomly, we can ensure that any bias in the sample is due to random chance. For our initial purposes, we work within the context of simple random sampling. We must decide how large of a sample to use; we denote the sample size as \(n\). The random experiment is defined as randomly selecting \(n\) objects from the population. The sample space of our random experiment consists of all the possible samples of size \(n\) taken from the population of size \(N.\) There are \(\sideset{_{N}}{_{n}}C\) possible samples. Each sample is assigned a value by computing the sample statistic of interest. These possible values, along with their probabilities, form the probability distribution of the sample statistic under simple random sampling. The questions of interest are: what values can the sample statistic take on, and what are the probabilities?
Constructing Probability Distributions of Sample Statistics: Proportions
We cannot know the values and their associated probabilities without studying the entire population. However, we can build solid intuition by studying small populations, where we can exhaustively study the probability distributions. We will be able to generalize in the following sections.
Consider the family of five that has been with us throughout the text. Adam and Betsy have three children: Cathy, Damon, and Erin. This family will serve as our initial population of interest. We will study several different characteristics of this family. The first step is to decide how many family members we want to sample. Indeed, it is unnecessary to sample in this situation, but this process builds our intuition and understanding of the topic. Sampling \(3\) family members will be perfect for our example.
- Determine the number of samples and then list all possible samples of size \(n=3\) from the population of Adam \((A)\), Betsy \((B)\), Cathy \((C)\), Damon \((D)\), and Erin \((E).\)
- Answer
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We are selecting \(3\) family members from a family of \(5\). The order in which we select them does not matter; therefore, there are \(\sideset{_{5}}{_{3}}C\) \(=10\) possible samples.\begin{array}{|ccccc|}\hline ABC&ABD&ABE&ACD&ACE \\ ADE&BCD&BCE&BDE&CDE \\ \hline\nonumber\end{array}Notice the pattern of our labelling. Adopting such a method ensures all possible samples are listed easily.
- Recall that Adam, Betsy, and Cathy all wear glasses. The population proportion \(p\) of family members that wear glasses is \(p=\frac{3}{5}\) \(=60\%\). For each possible sample of size \(3,\) determine the sample proportion \(\hat{p}\).
- Answer
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We must compute the sample proportion for each of the \(10\) samples from part one of this text exercise.
Table \(\PageIndex{1}\): All possible samples and their sample proportions
Sample \(\hat{p}\) Sample \(\hat{p}\) \(ABC\) \(\dfrac{3}{3}=1\) \(ADE\) \(\dfrac{1}{3}\) \(ABD\) \(\dfrac{2}{3}\) \(BCD\) \(\dfrac{2}{3}\) \(ABE\) \(\dfrac{2}{3}\) \(BCE\) \(\dfrac{2}{3}\) \(ACD\) \(\dfrac{2}{3}\) \(BDE\) \(\dfrac{1}{3}\) \(ACE\) \(\dfrac{2}{3}\) \(CDE\) \(\dfrac{1}{3}\) Note that none of the possible sample proportion values equal the population proportion. Recall from our first text exercise with this family that this is true for all possible sample sizes in this particular context. Still, it is not necessarily true for others (see the last part of the referenced exercise for a refresher).
- We are considering all the samples of size \(3\) and their sample proportions and have not conducted a random sampling to produce one of these possibilities. We are developing our understanding of the sample proportion \(\hat{p}\) as a random variable. There are three possible values that \(\hat{p}\) takes on: \(\frac{1}{3},\) \(\frac{2}{3},\) and \(1.\) Our task now is to fully understand \(\hat{p}\) as a random variable and construct its probability distribution.
- Answer
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Since we are conducting a simple random sampling of size \(3\) from the family of \(5\), each sample is equally probable. We can determine the probabilities of our random variable \(\hat{p}\) using the classical approach to probability.
Table \(\PageIndex{2}\): Probability distribution of sample proportions
\(\hat{p}\) \(P(\hat{p})\) \(\dfrac{1}{3}\) \(\dfrac{3}{10}\) \(\dfrac{2}{3}\) \(\dfrac{6}{10}=\dfrac{3}{5}\) \(1\) \(\dfrac{1}{10}\)
- Determine the expected value, variance, and standard deviation of our random variable \(\hat{p}.\)
- Answer
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Table \(\PageIndex{3}\): Computation table
\(\hat{p}\) \(P(\hat{p})\) \(\hat{p}\cdot P(\hat{p})\) \(\left(\hat{p}-\mu_{\hat{p}}\right)^2\cdot P(\hat{p})\) \(\dfrac{1}{3}\) \(\dfrac{3}{10}\) \(\dfrac{1}{3}\cdot\dfrac{3}{10}=\dfrac{1}{10}\) \( \left(\dfrac{1}{3}-\dfrac{3}{5}\right)^2 \cdot \dfrac{3}{10}=\dfrac{48}{2250}\) \(\dfrac{2}{3}\) \(\dfrac{6}{10}=\dfrac{3}{5}\) \(\dfrac{2}{3}\cdot\dfrac{6}{10}=\dfrac{4}{10}=\dfrac{2}{5}\) \( \left(\dfrac{2}{3}-\dfrac{3}{5}\right))^2 \cdot \dfrac{3}{5}=\dfrac{3}{1125}\) \(1\) \(\dfrac{1}{10}\) \(1\cdot\dfrac{1}{10}=\dfrac{1}{10}\) \( \left(1-\dfrac{3}{5}\right))^2 \cdot \dfrac{1}{10}=\dfrac{2}{125}\) \(\mu_{\hat{p}}=\text{E}(\hat{p})=\dfrac{1}{10}+\dfrac{4}{10}+\dfrac{1}{10}=\dfrac{6}{10}=\dfrac{3}{5}\) \(\sigma^2_{\hat{p}}=\text{Var}(\hat{p})=\dfrac{48}{2250}+\dfrac{6}{2250}+\dfrac{36}{2250}=\dfrac{90}{2250}=\dfrac{1}{25}\) \(\sigma_{\hat{p}}=\sqrt{\text{Var}(\hat{p})}=\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\) Note that \(\mu_{\hat{p}}=\frac{3}{5}=p.\) even though none of the possible sample proportions are equal to the population proportion, the expected value of the probability distribution of sample proportions is the population proportion.
When considering probability distributions of sample statistics (for example, sample proportions in our previous exercise), we use subscripts to indicate the sample statistic. This visual reminder provides a simple method of improving clarity and reducing the risk of errors. Reading symbolic expressions with meaning is an important skill to maintain and develop, especially at this point in the course.
Constructing Probability Distributions of Sample Statistics: Means
To continue our exploration, we consider additional data regarding our family of five (presented below): the number of states visited by each person. When considering the characteristic of needing eyeglasses, a qualitative variable, each member of our population either possessed the characteristic or did not. Proportions and modes would be natural sample statistics to consider. On the other hand, the number of states visited is a quantitative variable, so we have many more options to consider for sample statistics. We shall consider some in the following text exercises.
Table \(\PageIndex{4}\): Number of States Visited
| Family Member | Number of States Visited |
|---|---|
| Adam | \(20\) |
| Betsy | \(30\) |
| Cathy | \(15\) |
| Damon | \(12\) |
| Erin | \(5\) |
- Construct the probability distribution of sample means using a sample size of \(3.\)
- Answer
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For each of the \(10\) samples, we must compute the sample mean. We do so by filling out a table.
Table \(\PageIndex{5}\): All possible samples and their sample means
Sample Sample Data \(\bar{x}\) Sample Sample Data \(\bar{x}\) \(ABC\) \(20, 30, 15\) \(\dfrac{20+30+15}{3}=\dfrac{65}{3}\) \(ADE\) \(20, 12, 5\) \(\dfrac{20+12+5}{3}=\dfrac{37}{3}\) \(ABD\) \(20, 30, 12\) \(\dfrac{20+30+12}{3}=\dfrac{62}{3}\) \(BCD\) \(30, 15, 12\) \(\dfrac{30+15+12}{3}=\dfrac{57}{3}=19\) \(ABE\) \(20, 30, 5\) \(\dfrac{20+30+5}{3}=\dfrac{55}{3}\) \(BCE\) \(30, 15, 5\) \(\dfrac{30+15+5}{3}=\dfrac{50}{3}\) \(ACD\) \(20, 15, 12\) \(\dfrac{20+15+12}{3}=\dfrac{47}{3}\) \(BDE\) \(30, 12, 5\) \(\dfrac{30+12+5}{3}=\dfrac{47}{3}\) \(ACE\) \(20, 15, 5\) \(\dfrac{20+15+5}{3}=\dfrac{40}{3}\) \(CDE\) \(15, 12, 5\) \(\dfrac{15+12+5}{3}=\dfrac{32}{3}\) Table \(\PageIndex{6}\): Probability distribution of sample means
\(\bar{x}\) \(P(\bar{x})\) \(\dfrac{32}{3}\) \(\dfrac{1}{10}\) \(\dfrac{37}{3}\) \(\dfrac{1}{10}\) \(\dfrac{40}{3}\) \(\dfrac{1}{10}\) \(\dfrac{47}{3}\) \(\dfrac{2}{10}=\dfrac{1}{5}\) \(\dfrac{50}{3}\) \(\dfrac{1}{10}\) \(\dfrac{55}{3}\) \(\dfrac{1}{10}\) \(\dfrac{57}{3}\) \(\dfrac{1}{10}\) \(\dfrac{62}{3}\) \(\dfrac{1}{10}\) \(\dfrac{65}{3}\) \(\dfrac{1}{10}\)
- Determine the expected value, variance, and standard deviation of our random variable \(\bar{x}\). Compare the expected value of the probability distribution \(\mu_{\bar{x}}\) with population mean \(\mu\).
- Answer
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Table \(\PageIndex{7}\): Table of computations
\(\bar{x}\) \(P(\bar{x})\) \(\bar{x}\cdot P(\bar{x})\) \(\left(\bar{x}-\mu_{\bar{x}}\right)^2\cdot P(\bar{x})\) \(\dfrac{32}{3}\) \(\dfrac{1}{10}\) \(\dfrac{32}{3}\cdot\dfrac{1}{10}=\dfrac{32}{30}\) \(\left(\dfrac{32}{3}-\dfrac{82}{5} \right)^2 \cdot \dfrac{1}{10}=\dfrac{7396}{2250}\) \(\dfrac{37}{3}\) \(\dfrac{1}{10}\) \(\dfrac{37}{3}\cdot\dfrac{1}{10}=\dfrac{37}{30}\) \(\left(\dfrac{37}{3}-\dfrac{82}{5} \right)^2 \cdot \dfrac{1}{10}=\dfrac{3721}{2250}\) \(\dfrac{40}{3}\) \(\dfrac{1}{10}\) \(\dfrac{40}{3}\cdot\dfrac{1}{10}=\dfrac{40}{30}\) \(\left(\dfrac{40}{3}-\dfrac{82}{5}\right)^2 \cdot \dfrac{1}{10}=\dfrac{2116}{2250}\) \(\dfrac{47}{3}\) \(\dfrac{2}{10}=\dfrac{1}{5}\) \(\dfrac{47}{3}\cdot\dfrac{2}{10}=\dfrac{94}{30}\) \(\left(\dfrac{47}{3}-\dfrac{82}{5}\right)^2 \cdot \dfrac{2}{10}=\dfrac{242}{2250}\) \(\dfrac{50}{3}\) \(\dfrac{1}{10}\) \(\dfrac{50}{3}\cdot\dfrac{1}{10}=\dfrac{50}{30}\) \(\left(\dfrac{50}{3}-\dfrac{82}{5}\right)^2 \cdot \dfrac{1}{10}=\dfrac{16}{2250}\) \(\dfrac{55}{3}\) \(\dfrac{1}{10}\) \(\dfrac{55}{3}\cdot\dfrac{1}{10}=\dfrac{55}{30}\) \(\left(\dfrac{55}{3}-\dfrac{82}{5}\right)^2 \cdot \dfrac{1}{10}=\dfrac{841}{2250}\) \(\dfrac{57}{3}\) \(\dfrac{1}{10}\) \(\dfrac{57}{3}\cdot\dfrac{1}{10}=\dfrac{57}{30}\) \(\left(\dfrac{57}{3}-\dfrac{82}{5}\right)^2 \cdot \dfrac{1}{10}=\dfrac{1521}{2250}\) \(\dfrac{62}{3}\) \(\dfrac{1}{10}\) \(\dfrac{62}{3}\cdot\dfrac{1}{10}=\dfrac{62}{30}\) \(\left(\dfrac{62}{3}-\dfrac{82}{5}\right)^2 \cdot \dfrac{1}{10}=\dfrac{4096}{2250}\) \(\dfrac{65}{3}\) \(\dfrac{1}{10}\) \(\dfrac{65}{3}\cdot\dfrac{1}{10}=\dfrac{65}{30}\) \(\left(\dfrac{65}{3}-\dfrac{82}{5}\right)^2 \cdot \dfrac{1}{10}=\dfrac{6241}{2250}\) \(\mu_{\bar{x}}=\text{E}(\bar{x})=\dfrac{32}{30}+\dfrac{37}{30}+\ldots+\dfrac{65}{30}=\dfrac{492}{30}=\dfrac{82}{5}=16.4\) \(\sigma^2_{\bar{x}}=\text{Var}(\bar{x})=\dfrac{7396}{2250}+\dfrac{3721}{2250}+\ldots+\dfrac{6241}{2250}=\dfrac{26,190}{2250}=\dfrac{291}{25}=11.64\) \(\sigma_{\bar{x}}=\sqrt{\text{Var}(\bar{x})}=\sqrt{\dfrac{291}{25}}\approx3.4117\) To compute the population mean \(\mu,\) we have \(\frac{20+30+15+12+5}{5}\) \(=\frac{82}{5}\) \(=16.4\) meaning \(\mu_{\bar{x}}=\mu.\) The average of all the sample means is the same as the population mean.
Constructing Probability Distributions of Sample Statistics: Range
- Construct the probability distribution of sample ranges using a sample size of \(3.\)
- Answer
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For each of the \(10\) samples, we must compute the sample range \(\text{range}.\) We again do so by filling out a table.
Table \(\PageIndex{8}\): All possible samples and their sample ranges
Sample Sample Data Sample Range \((\text{range})\) Sample Sample Data Sample Range \((\text{range})\) \(ABC\) \(20, 30, 15\) \(15\) \(ADE\) \(20, 12, 5\) \(15\) \(ABD\) \(20, 30, 12\) \(18\) \(BCD\) \(30, 15, 12\) \(18\) \(ABE\) \(20, 30, 5\) \(25\) \(BCE\) \(30, 15, 5\) \(25\) \(ACD\) \(20, 15, 12\) \(8\) \(BDE\) \(30, 12, 5\) \(25\) \(ACE) \(20, 15, 5\) \(15\) \(CDE\) \(15, 12, 5\) \(10\) Table \(\PageIndex{9}\): Probability distribution of sample ranges
\(\text{range}\) \(P(\text{range})\) \(8\) \(\dfrac{1}{10}\) \(10\) \(\dfrac{1}{10}\) \(15\) \(\dfrac{3}{10}\) \(18\) \(\dfrac{2}{10}=\dfrac{1}{5}\) \(25\) \(\dfrac{3}{10}\)
- Determine our random variable's expected value, variance, and standard deviation \(\text{range}\). Compare the expected value of the probability distribution \(\mu_{\bar{x}}\) with the population range.
- Answer
-
Table \(\PageIndex{10}\): Table of computations
\(\text{range}\) \(P(\text{range})\) \(\text{range}\cdot P(\text{range})\) \(\left(\text{range}-\mu_{\text{range}}\right)^2\cdot P(\text{range})\) \(8\) \(\dfrac{1}{10}\) \(8\cdot\dfrac{1}{10}=\dfrac{8}{10}=\dfrac{4}{5}\) \(\left(8-\dfrac{87}{5}\right)^2 \cdot \dfrac{1}{10}=\dfrac{2209}{250}\) \(10\) \(\dfrac{1}{10}\) \(10\cdot\dfrac{1}{10}=\dfrac{10}{10}=1\) \(\left(10-\dfrac{87}{5}\right)^2 \cdot \dfrac{1}{10}=\dfrac{1369}{250}\) \(15\) \(\dfrac{3}{10}\) \(15\cdot\dfrac{3}{10}=\dfrac{45}{10}=\dfrac{9}{2}\) \(\left(15-\dfrac{87}{5}\right)^2 \cdot \dfrac{3}{10}=\dfrac{432}{250}\) \(18\) \(\dfrac{2}{10}=\dfrac{1}{5}\) \(18\cdot\dfrac{2}{10}=\dfrac{36}{10}=\dfrac{18}{5}\) \(\left(18-\dfrac{87}{5}\right)^2 \cdot \dfrac{2}{10}=\dfrac{18}{250}\) \(25\) \(\dfrac{3}{10}\) \(25\cdot\dfrac{3}{10}=\dfrac{75}{10}=\dfrac{15}{2}\) \(\left(25-\dfrac{87}{5}\right)^2 \cdot \dfrac{3}{10}=\dfrac{4332}{250}\) \(\mu_{\text{range}}=\text{E}(\text{range})=\dfrac{4}{5}+1+\ldots+\dfrac{15}{2}=\dfrac{174}{10}=\dfrac{87}{5}=17.4\) \(\sigma^2_{\text{range}}=\text{Var}(\text{range})=\dfrac{2209}{250}+\dfrac{1369}{250}+\ldots+\dfrac{4332}{250}=\dfrac{8360}{250}=\dfrac{836}{25}=33.44\) \(\sigma_{\text{range}}=\sqrt{\text{Var}(\text{range})}=\sqrt{\dfrac{207}{10}}\approx5.7827\) To compute the population range, we have \(30-5\) \(=25\) meaning \(\mu_{\text{range}}\ne \text{Population Range}.\) The average of the sample ranges is not the same as the population range. We might have expected this. A sample range could equal the population range if it includes the largest and smallest values. However, for most samples, the sample range will be smaller than the population range because the largest and smallest data values will not be included in the sample. Moreover, the sample range can never exceed the population range. Therefore, the average of the sample ranges is an average of numbers that are never larger than the population range, making the average smaller. This reasoning implies that for any sufficiently varied population, the average of the sample ranges will be less than the population range. Contrast this with the concept of a sample mean.
Probability Distributions of Sample Statistics and Sample Size
In each of our previous examples, we used the sample size \(n=3.\) We could have sampled a single person \((n=1)\) or any number less than \(5.\) If \(n\) were to be \(5\) in our context, \(n=N,\) we would be studying the entire population, not sampling from it, and then computing the parameter rather than estimating it. We now explore what happens to the probability distributions as we change the sample size while remaining in the same population. Again, we will consider the proportion of family members who wear glasses. The probability distribution of sample proportions for \(n=1,2,3,\) and \(4\) and the expected value, variance, and standard deviation are reported in the table below.
Table \(\PageIndex{11}\): Probability Distribution of Sample Proportions for \(n=1,2,3,\) and \(4\)
| \(n=1\) | \(n=2\) | \(n=3\) | \(n=4\) | |
|---|---|---|---|---|
| \(\mu_{\hat{p}}\) | \(\dfrac{3}{5}\) | \(\dfrac{3}{5}\) | \(\dfrac{3}{5}\) | \(\dfrac{3}{5}\) |
| \(\sigma^2_{\hat{p}}\) | \(0.24\) | \(0.09\) | \(0.04\) | \(0.015\) |
| \(\sigma_{\hat{p}}\) | \(\approx0.4899\) | \(0.3\) | \(0.2\) | \(\approx0.1225\) |
- Students struggling to understand these probability distributions and their construction are encouraged to verify the results above by first building each probability distribution and then computing each measure . Click to check your work.
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- Probability distribution of sample proportions \(n=1\)
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Table \(\PageIndex{12}\): Probability Distribution of Sample Proportions \(n=1\) (\(5\) Samples)
\(n=1\) (\(5\) Samples) \(\hat{p}\) \(P(\hat{p})\) \(0\) \(\dfrac{2}{5}\) \(1\) \(\dfrac{3}{5}\) \(\color{white}\dfrac{1}{1}\) \(\mu_{\hat{p}}=\dfrac{3}{5}\) \(\sigma^2_{\hat{p}}=0.24\) \(\sigma_{\hat{p}}\approx0.4899\)
- Probability distribution of sample proportions \(n=2\)
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Table \(\PageIndex{13}\): Probability Distribution of Sample Proportions \(n=2\) (\(10\) Samples)
\(n=2\) (\(10\) Samples) \(\hat{p}\) \(P(\hat{p})\) \(0\) \(\dfrac{1}{10}\) \(\dfrac{1}{2}\) \(\dfrac{6}{10}\) \(1\) \(\dfrac{3}{10}\) \(\mu_{\hat{p}}=\dfrac{3}{5}\) \(\sigma^2_{\hat{p}}=0.09\) \(\sigma_{\hat{p}}=0.3\)
- Probability distribution of sample proportions \(n=3\)
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Table \(\PageIndex{14}\): Probability Distribution of Sample Proportions \(n=3\) (\(10\) Samples)
\(n=3\) (\(10\) Samples) \(\hat{p}\) \(P(\hat{p})\) \(\dfrac{1}{3}\) \(\dfrac{3}{10}\) \(\dfrac{2}{3}\) \(\dfrac{6}{10}=\dfrac{3}{5}\) \(1\) \(\dfrac{1}{10}\) \(\mu_{\hat{p}}=\dfrac{3}{5}\) \(\sigma^2_{\hat{p}}=0.04\) \(\sigma_{\hat{p}}=0.2\)
- Probability distribution of sample proportions \(n=4\)
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Table \(\PageIndex{15}\): Probability Distribution of Sample Proportions \(n=4\) (\(5\) Samples)
\(n=4\) (\(5\) Samples) \(\hat{p}\) \(P(\hat{p})\) \(\dfrac{1}{2}\) \(\dfrac{6}{10}\) \(\dfrac{3}{4}\) \(\dfrac{4}{10}\) \(\color{white}\dfrac{1}{1}\) \(\mu_{\hat{p}}=\dfrac{3}{5}\) \(\sigma^2_{\hat{p}}=0.015\) \(\sigma_{\hat{p}}\approx0.1225\)
The expected value of each probability distribution of sample proportions is the same as the population proportion, regardless of the sample size; however, the variance and standard deviation values change with the sample size. The variance decreases as \(n\) increases, indicating that as \(n\) increases, the probability distribution is packed more closely around the population proportion. The spread decreases while remaining centered on the population proportion. Let us construct the probability distributions of sample means and ranges to see if similar patterns emerge.
Table \(\PageIndex{16}\): Probability Distribution of Sample Means \(n=1,2,3,\) and \(4\)
| \(n=1\) | \(n=2\) | \(n=3\) | \(n=4\) | |
|---|---|---|---|---|
| \(\mu_{\bar{x}}\) | \(\dfrac{3}{5}\) | \(\dfrac{3}{5}\) | \(\dfrac{3}{5}\) | \(\dfrac{3}{5}\) |
| \(\sigma^2_{\bar{x}}\) | \(0.24\) | \(0.09\) | \(0.04\) | \(0.015\) |
| \(\sigma_{\bar{x}}\) | \(\approx0.4899\) | \(0.3\) | \(0.2\) | \(\approx0.1225\) |
- Again, we encourage students struggling to understand these probability distributions and their construction to verify the results above by first building each probability distribution and then computing each measure. Click to check your work.
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- Probability distribution of sample means \(n=1\)
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Table \(\PageIndex{17}\): Probability Distribution of Sample Means \(n=1\) (\(5\) Samples)
\(n=1\) (\(5\) Samples) \(\bar{x}\) \(P(\bar{x})\) \(5\) \(\dfrac{1}{5}\) \(12\) \(\dfrac{1}{5}\) \(15\) \(\dfrac{1}{5}\) \(20\) \(\dfrac{1}{5}\) \(30\) \(\dfrac{1}{5}\) \(\mu_{\bar{x}}=16.4\) \(\sigma^2_{\bar{x}}=69.84\) \(\sigma_{\bar{x}}\approx8.3570\)
- Probability distribution of sample means \(n=2\)
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Table \(\PageIndex{18}\): Probability Distribution of Sample Means \(n=2\) (\(10\) Samples)
\(n=2\) (\(10\) Samples) \(\bar{x}\) \(P(\bar{x})\) \(8.5\) \(\dfrac{1}{10}\) \(10\) \(\dfrac{1}{10}\) \(12.5\) \(\dfrac{1}{10}\) \(13.5\) \(\dfrac{1}{10}\) \(16\) \(\dfrac{1}{10}\) \(17.5\) \(\dfrac{2}{10}\), \(21\) \(\dfrac{1}{10}\) \(22.5\) \(\dfrac{1}{10}\) \(25\) \(\dfrac{1}{10}\) \(\mu_{\bar{x}}=16.4\) \(\sigma^2_{\bar{x}}=26.19\) \(\sigma_{\bar{x}}\approx5.1176\)
- Probability distribution of sample means \(n=3\)
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Table \(\PageIndex{19}\): Probability Distribution of Sample Means \(n=3) (\(10\) Samples)
\(n=3\) (\(10\) Samples) \(\bar{x}\) \(P(\bar{x})\) \(10.\bar{6}\) \(\dfrac{1}{10}\) \(12.\bar{3}\) \(\dfrac{1}{10}\) \(13.\bar{3}\) \(\dfrac{1}{10}\) \(15.\bar{6}\) \(\dfrac{2}{10}\) \(16.\bar{6}\) \(\dfrac{1}{10}\) \(18.\bar{3}\) \(\dfrac{1}{10}\) \(19\) \(\dfrac{1}{10}\) \(20.\bar{6}\) \(\dfrac{1}{10}\) \(21.\bar{6}\) \(\dfrac{1}{10}\) \(\mu_{\bar{x}}=16.4\) \(\sigma^2_{\bar{x}}=11.64\) \(\sigma_{\bar{x}}\approx3.4117\)
- Probability distribution of sample means \(n=4\)
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Table \(\PageIndex{20}\): Probability Distribution of Sample Means \(n=4\) (\(5\) Samples)
\(n=4\) (\(5\) Samples) \(\bar{x}\) \(P(\bar{x})\) \(13\) \(\dfrac{1}{5}\) \(13.5\) \(\dfrac{1}{5}\) \(16.75\) \(\dfrac{1}{5}\) \(17.5\) \(\dfrac{1}{5}\) \(19.25\) \(\dfrac{1}{5}\) \(\mu_{\bar{x}}=16.4\) \(\sigma^2_{\bar{x}}=4.365\) \(\sigma_{\bar{x}}\approx2.0893\)
We notice a similar trend with the probability distributions of sample means. Regardless of our sample size \(n,\) the expected value is the population mean. As \(n\) increases, the spread of our distributions decreases. We now look at a final example: sample ranges. Range, indeed all measures of spread, are not very informative (nor well-defined) if there is only one observation. Think about why that is. We will only consider samples with at least \(2\) observations.
Table \(\PageIndex{21}\): Probability Distribution of Sample Ranges \(n=1,2,3,\) and \(4\)
| \(n=2\) | \(n=3\) | \(n=4\) | |
|---|---|---|---|
| \(\mu_{\text{range}}\) | \(11.6\) | \(17.4\) | \(21.6\) |
| \(\sigma^2_{\text{range}}\) | \(40.04\) | \(33.44\) | \(18.24\) |
| \(\sigma_{\text{range}}\) | \(\approx6.3277\) | \(\approx5.7827\) | \(\approx4.2708\) |
- Click to verify probability distribution construction as needed.
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- Probability distribution of sample ranges \(n=2\)
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Table \(\PageIndex{22}\): Probability Distribution of Sample Ranges \(n=2\) (\(10\) Samples)
\(n=2\) (\(10\) Samples) \(\text{range}\) \(P(\text{range})\) \(3\) \(\dfrac{1}{10}\) \(5\) \(\dfrac{1}{10}\) \(7\) \(\dfrac{1}{10}\) \(8\) \(\dfrac{1}{10}\) \(10\) \(\dfrac{2}{10}\) \(15\) \(\dfrac{2}{10}\) \(18\) \(\dfrac{1}{10}\) \(25\) \(\dfrac{1}{10}\) \(\mu_{\text{range}}=11.6\) \(\sigma^2_{\text{range}}=40.04\) \(\sigma_{\text{range}}\approx6.3277\)
- Probability distribution of sample ranges \(n=3\)
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Table \(\PageIndex{23}\): Probability Distribution of Sample Ranges \(n=3\) (\(10\) Samples)
\(n=3\) (\(10\) Samples) \(\text{range}\) \(P(\text{range})\) \(8\) \(\dfrac{1}{10}\) \(10\) \(\dfrac{1}{10}\) \(15\) \(\dfrac{1}{10}\) \(18\) \(\dfrac{2}{10}\) \(25\) \(\dfrac{1}{10}\) \(\mu_{\text{range}}=17.4\) \(\sigma^2_{\text{range}}=33.44\) \(\sigma_{\text{range}}\approx5.7827\)
- Probability distribution of sample ranges \(n=4\)
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Table \(\PageIndex{24}\): Probability Distribution of Sample Ranges \(n=4\) (\(5\) Samples)
\(n=4\) (\(5\) Samples) \(\text{range}\) \(P(\text{range})\) \(15\) \(\dfrac{1}{5}\) \(18\) \(\dfrac{1}{5}\) \(25\) \(\dfrac{3}{5}\) \(\mu_{\text{range}}=21.6\) \(\sigma^2_{\text{range}}=18.24\) \(\sigma_{\text{range}}\approx4.2708\)
Having seen that the expected value of the probability distribution of sample ranges was not the population range in the case of \(n=3,\) it is not surprising to see a similar situation in the other two sample sizes. The expected value of the probability distribution equaling the population parameter is not typical among the various statistics we consider. However, as \(n\) increases, we see a decrease in the spread of our probability distributions.
Sampling Distributions of Sample Statistics
Let us review what we know so far. We have constructed probability distributions of sample statistics under simple random sampling by computing a particular sample statistic for every possible sample of a specific size, \(n,\) and then determining the probability that these values occur. For some sample statistics, the probability distribution is centered around the population parameter; that is, for some statistics, the expected value of the probability distribution is the associated population parameter. Finally, we noticed that the spread of the probability distribution decreases as the sample size increases for each of the statistics studied so far.
We now introduce the probability distribution that much of inferential statistics is built upon, the sampling distribution of sample statistics . A sampling distribution is similar in nature to the probability distributions that we have been building in this section, but with one fundamental difference: rather than sampling using simple random sampling, the sampling method is to select randomly \(n\) objects, one at a time, from the population with replacement . Note that the order of selection will matter. As such, when considering a population of size \(N,\) there are \(N\) possibilities for each random selection, indicating that there are \(N^n\) samples to consider instead of the \(\sideset{_{N}}{_{n}}C\) as with simple random sampling.
One may question why a distribution constructed from sampling with replacement takes priority in inferential statistics when the probability distributions above seem much more intuitive and easy to construct (having less samples to consider). As it so happens, when populations are large enough compared to the sample size (we will discuss this more later), the probability distributions of sample statistics constructed from simple random sampling are approximated well using the sampling distribution of sample statistics.
What intuition we have built from our previous constructions transfer directly to sampling distributions. As the sample size increases, the spread of the sampling distribution decreases. This should appeal to our basic intuition that larger samples better represent the population than smaller samples. For certain statistics, the expected value of the sampling distribution is the population parameter. We call such sample statistics unbiased estimators. Most introductory statistics books examine three unbiased estimators: sample means, proportions, and variances. The sample range is a biased estimator of the population range for the same reasons discussed above.
- Within the context of the family of five from above and using a sample size of \(3,\) construct the sampling distribution of sample proportions for the proportion of family members who wear glasses. Recall that the sampling method used in producing sampling distributions is selecting a member at random \(n\) times with replacement which means that order matters and the same member could be in the sample multiple times.
- Answer
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We begin our solution by counting the number of samples that we need to consider. Since the population size is \(N=5\) and our sample size is \(n=3,\) we have \(5^3\) \(=125\) different possible samples. That is a lot!
\begin{array}{|ccccc|}\hline AAA&BAA&CAA&DAA&EAA \\ AAB&BAB&CAB&DAB&EAB \\ AAC&BAC&CAC&DAC&EAC \\ AAD&BAD&CAD&DAD&EAD \\ AAE&BAE&CAE&DAE&EAE \\ ABA&BBA&CBA&DBA&EBA \\ ABB&BBB&CBB&DBB&EBB \\ ABC&BBC&CBC&DBC&EBC \\ ABD&BBD&CBD&DBD&EBD \\ ABE&BBE&CBE&DBE&EBE \\ ACA&BCA&CCA&DCA&ECA \\ ACB&BCB&CCB&DCB&ECB \\ ACC&BCC&CCC&DCC&ECC \\ ACD&BCD&CCD&DCD&ECD \\ ACE&BCE&CCE&DCE&ECE \\ ADA&BDA&CDA&DDA&EDA \\ ADB&BDB&CDB&DDB&EDB \\ ADC&BDC&CDC&DDC&EDC \\ ADD&BDD&CDD&DDD&EDD \\ ADE&BDE&CDE&DDE&EDE \\ AEA&BEA&CEA&DEA&EEA \\ AEB&BEB&CEB&DEB&EEB \\ AEC&BEC&CEC&DEC&EEC \\ AED&BED&CED&DED&EED \\ AEE&BEE&CEE&DEE&EEE \\ \hline\nonumber\end{array}
We leave the computation of each sample proportion to the reader and tabulate the results in the following table.
Table \(\PageIndex{25}\): Sampling distribution of sample proportions
\(\hat{p}\) \(P(\hat{p})\) \(0\) \(\dfrac{8}{125}\) \(\dfrac{1}{3}\) \(\dfrac{36}{125}\) \(\dfrac{2}{3}\) \(\dfrac{54}{125}\) \(1\) \(\dfrac{27}{125}\)
- Compute the expected value, variance, and standard deviation of the sampling distribution of sample proportions found in the previous portion of this text exercise.
- Answer
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Table \(\PageIndex{26}\): Table of computations
\(\hat{p}\) \(P(\hat{p})\) \(\hat{p}\cdot P(\hat{p})\) \(\left(\hat{p}-\mu_{\hat{p}}\right)^2\cdot P(\hat{p})\) \(0\) \(\dfrac{8}{125}\) \(0\) \( \left(0-\dfrac{3}{5}\right)^2 \cdot \dfrac{8}{125}=\dfrac{72}{3125}\) \(\dfrac{1}{3}\) \(\dfrac{36}{125}\) \(\dfrac{1}{3}\cdot\dfrac{36}{125}=\dfrac{12}{125}\) \( \left(\dfrac{1}{3}-\dfrac{3}{5}\right)^2 \cdot \dfrac{36}{125}=\dfrac{576}{28125}=\dfrac{64}{3125}\) \(\dfrac{2}{3}\) \(\dfrac{54}{125}\) \(\dfrac{2}{3}\cdot\dfrac{54}{125}=\dfrac{36}{125}\) \( \left(\dfrac{2}{3}-\dfrac{3}{5}\right))^2 \cdot \dfrac{54}{125}=\dfrac{54}{28125}=\dfrac{6}{3125}\) \(1\) \(\dfrac{27}{125}\) \(1\cdot\dfrac{27}{125}=\dfrac{27}{125}\) \( \left(1-\dfrac{3}{5}\right))^2 \cdot \dfrac{27}{125}=\dfrac{108}{3125}\) \(\mu_{\hat{p}}=\text{E}(\hat{p})=\dfrac{12}{125}+\dfrac{36}{125}+\dfrac{27}{125}=\dfrac{75}{125}=\dfrac{3}{5}\) \(\sigma^2_{\hat{p}}=\text{Var}(\hat{p})=\dfrac{72}{3125}+\dfrac{64}{3125}+\dfrac{6}{3125}+\dfrac{108}{3125}=\dfrac{250}{3125}=\dfrac{2}{25}\) \(\sigma_{\hat{p}}=\sqrt{\text{Var}(\hat{p})}=\sqrt{\dfrac{2}{25}}\approx0.2828\)
Sampling Distributions and Large Populations
The number of possible samples is quite large when we study most populations. As an example, FHSU enrolls about \(3600\) on-campus students in a typical semester. We might consider randomly sampling \(30\) students. In which case, the number of possible samples is about \(4.8874\cdot10^{106}.\) We naturally expect many possible sample statistic values with so many different possible samples. As such, we expect that sampling distributions take on large numbers of values and can be reasonably represented as continuous random variables, which can be understood using histograms and probability density functions.
With larger populations, we can no longer expect to construct all possible samples of a given size, \(n,\) from a population to develop an understanding of sampling distributions. However, that does not mean we cannot build a reasonably accurate representation. We can approximate sampling distributions by randomly sampling from all the possible samples and then constructing histograms to visualize the shape of the distribution. We build relative frequency histograms to estimate the probability distribution of the sampling distribution. This process is tedious but can be easily implemented with a computer.
The Online StatBook Project provides a program that operates with frequency counts rather than relative frequency counts. Since both preserve the general shape of a distribution, we can build an intuition about sampling distributions. Open the link and read the instructions page carefully. Then click on the "Begin" button at the top left part of your screen. Note: within this program, \(N\) stands for sample size, not population size.
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Use the following settings for this program.
Parent Population:
Normal.
First Sampling Distribution:
Mean \(N=5\).
Second
Sampling Distribution:
None \(N=5\). Click the "Animated" button and watch the animation.
- What do the boxes in the second distribution represent?
- What does the singular box in the third distribution mean?
- Construct a sampling distribution using \(125\) random samples (Reps=\(125\)), then \(10,125\), and \(50,125\). What is happening as we randomly sample more and more?
- Answer
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The five boxes in the second distribution represent the five observations randomly selected to form our single sample of size \(5\). The singular box in the third distribution is the sample mean of the sample in the second distribution. The distributions should look different since we are running a simulation randomly selecting samples, but they should be reasonably close to the figure provided below.
Figure \(\PageIndex{1}\): Sampling distribution simulation
As we take more and more samples of size \(5\) from the population, we eventually converge to a consistent shape. Once the distribution does not change much, we have a decent idea of the sampling distribution.
- Use the following settings for this program. Parent Population: Normal. First Sampling Distribution: Mean \(N=2\) Fit Normal Checked. Second Sampling Distribution: Mean \(N=5\) Fit Normal Checked. Run the simulation using \(100,000\) random samples to estimate the sampling distributions of sample means. Compare the two sampling distributions.
- Answer
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Once again, our distributions will not be the same as those produced by the simulation for you, but they should be quite similar.
Figure \(\PageIndex{2}\): Sampling distribution simulation
Both sampling distributions are centered reasonably close to the population mean. When the sample size is \(5\), the sampling distribution is less spread out compared to the sampling distribution of sample size \(2\). Both sampling distributions have less spread than the parent population and fit the normal curve well.
- Use the following settings for this program. Parent Population: Skewed. First Sampling Distribution: Mean \(N=2\) Fit Normal Checked. Second Sampling Distribution: Mean \(N=20\) Fit Normal Checked. Run the simulation using \(100,000\) random samples to estimate the sampling distributions of sample means. Compare the two sampling distributions.
- Answer
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Figure \(\PageIndex{3}\): Sampling distribution simulation
Both sampling distributions are centered reasonably close to the population mean. The spread again decreases as the sample size increases. The first sampling distribution appears skewed to the right, just like the parent population, and is not normal. With a greater sample size, the second sampling distribution fits a normal curve much better.
- Use the following settings for this program. Parent Population: Custom. First Sampling Distribution: Mean \(N=10\) Fit Normal Checked. Second Sampling Distribution: Mean \(N=25\) Fit Normal Checked. Using your mouse, construct a parent population so that when you run the simulation using \(100,000\) random samples to estimate the sampling distributions of sample means, it does not appear to be normal.
- Answer
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Figure \(\PageIndex{4}\): Sampling distribution simulation
It takes an extreme parent population for the sampling distributions not to fit a normal curve with a sample size of \(25.\)
- Use the following settings for this program. Parent Population: Skewed. First Sampling Distribution: Var (U) \(N=2\) Fit Normal Checked. Second Sampling Distribution: Var (U) \(N=25\) Fit Normal Checked. Run the simulation using \(100,000\) random samples to estimate the sampling distributions of sample variances. Compare the two sampling distributions.
- Answer
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Figure \(\PageIndex{5}\): Sampling distribution simulation
Notice that the variance of the parent population \(6.22^2=38.6884\) is very close to the expected values (mean) of the sampling distributions. The sampling distributions are roughly centered on the population parameter. Recall that variance was the third unbiased estimator. Again, the spread of the sampling distributions decreases as the sample size increases. These sampling distributions, however, do not appear to be normally distributed. It seems closer when the sample size is \(25\), but it is still not a great fit.
- In the previous parts of this text exercise, the mean of the sampling distributions has been very close to the population parameter. We do want to provide an intuition about unbiased estimators. Use the following settings for this program. Parent Population: Skewed. First Sampling Distribution: Range \(N=2, 5, 10, 16, 20, 25\). Second Sampling Distribution: None \(N=5\). For each sample size, run the simulation using \(100,000\) random samples to estimate the sampling distributions of sample ranges. Describe what happens to the expected value of the sampling distribution of sample ranges (the mean of the second distribution) as the sample size increases. How is this different from all other sampling distributions within this text exercise?
- Answer
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Figure \(\PageIndex{6}\): Sampling distribution simulation
The population range is about \(28\). If we look at the expected values of the sampling distributions of sample ranges as the sample size increases, we see that the expected values increase each time. This is in contrast to what happened with the sampling distributions of sample means and variances. Regardless of the sample size, the sampling distributions of sample means and variances had expected values close to the population parameter. The distributions were centered on the population parameter. Some sample means were less than the population mean; some were more. There was no inherent bias in the estimation using sample means or sample variances. This is not the case for sample ranges. We are guaranteed that the sample range is less than or equal to the population range. Why do you think this is true? Because of this, the sampling distribution cannot be centered on the population range unless we have a trivial population, making the sample range a biased estimator.
Bridging Theory and Application
While simple random sampling and random selection with replacement are two fundamentally different approaches to sampling, when populations are large enough and the sample size is not too large relative to the population size, we consider the two methods approximately interchangable in regards to the probability distributions of sample statistics that are produced. That is, when the size conditions are met (discussed in future sections), we utilize the sampling distribution of sample statistics for a given sample size \(n\) to understand the probability of sample statistic values from simple random samples of that same size \(n.\) It is important to note that statistical analyses have been developed for cases where the size conditions are not met, but that these considerations are beyond the scope of this text. Interested readers are encouraged to continue taking more advanced statistics courses.


3.13: Guidelines for Successful Therapeutic Interactions
After the purpose of the therapeutic interaction has been established, the following guidelines assist in conducting a successful interaction:
Maintaining a Low-Authority Profile
The nurse must strive to maintain a low-authority profile at the beginning of the conversation. As the conversation progresses, the nurse can use more directive techniques to find out specific information. There are usually differences in age, sex, occupation, cultural background, moral and religious convictions between the nurse and the patient. These differences make it impossible for the nurse to fully understand the patient’s behavior and reactions. It is therefore, important for the nurse to understand and accept differences in patients’ cultures and beliefs. When in doubt, check with the patient. If trust is established, patient will be willing to teach the nurse.
Understandable Language
The nurse should determine the patient’s level of understanding and if necessary change the use of language, comments and questions. Using the terminology which the patient does not understand can also frighten the patient and make him/her think that he/she has a more serious problem than he/she originally wanted help for. At the same time, the patient could give incorrect information because due to confusion, he/she may give affirmative answers to questions about symptoms that he/she has not actually experienced [ 18 ]. Nurses should share their aims with patients before expecting them to participate in the interaction. They should understand that there is a mutual understanding of each other’s point of departure. In an assessment interview, the nurse can, for instance, say: “Mr Jones, I would like to give you information on how to lose weight so as to bring down you high blood pressure, but I first need to find out what you already know about the condition”. It is not only important that the patients understand what nurses expect from the conversation; it is also essential that nurses understand the patients and convey this understanding before they participate in the conversation. When providing emotional support, this understanding is often all that is necessary. For nurses to understand patients, they must encourage them to talk – not just about facts, but also about their feelings. The nurse must listen more than speak, both to what the patient is saying verbally and what is being said non-verbally. Having listened carefully, the nurse then concentrates and responds empathetically to the patients’ feelings. Only when the nurse has a reasonably complete understanding of the patient’s situation and has communicated this understanding, can she proceed to interventions, such as giving information or solving a problem.
Tailor the Message to the Totality of the Person
Saying something does not necessary mean that the message has been received and understood. It is the responsibility of the nurse to ensure that the person with whom he/she is conversing understands the message. To ensure this, the message has to be adapted to the language, culture and socio-economic status of the patient. The emotional or physical condition of patients may also make it difficult for them to receive long of complicated messages or even any message. There may also be other disturbances in the immediate environment for example, noise that can make the patient not to hear or understand the message. The message must also be adapted to the age of the patient [ 10 ].
Validate the Interpretation With the Patient
Validation means that you ask the patient whether your interpretation is correct or not. You therefore, ask him/her to confirm your understanding of what he/she said. Many misunderstandings arise because people interpret other people’s words without checking their interpretation. The nurse should try to eliminate misunderstandings in the conversations by checking meaning with the patient.
Active Listening
Active listening means concentrating all your senses and thoughts on the speaker. One can usually deduce whether a person is listening actively by looking at the following non-verbal indicators:
- Is the eye contact maintained with the person who is speaking?
- Are the body and face turned towards the speaker?
It is, of course, also clear from the verbal responses:
- Are there regular verbal responses, even if these consist only of encouraging sounds?
- Does the response indicate understanding, not only of the facts, but also of the feelings and the implications of the facts?
It is much easier to speak than to listen. Nurses are, in general, very active people, who want help b acting quickly. To ‘just listen’ without expressing opinions or offering advice is therefore, often not in their nature. Active listening is a valuable skill to acquire [ 10 , 17 , 18 ].
Evaluate Own Communication
In the interest of nurse–patient relationship, it is essential that they ascertain whether their communication has been successful. The following criteria can be used:
- Simplicity : Say what you want to say concisely and without using difficult or unfamiliar terms.
- Clarity : Say precisely what you want to say without digressing, and support your verbal message with non-verbal indicators.
- Relevance : Make sure that your message suits the situation, the time and the person you are speaking to.
- Adaptability : Adapt your response to the clues the patient that the patient gives you.
- Respect : Always show respect for the individuality and dignity of the person you are speaking to [ 17 ].


2.8: Family Health
Assessing the current and past health of a client’s family is important because diseases sometimes have a genetic component . These data provide you with a risk profile for inherited conditions. While it is important to understand the risk and likelihood of illness based on family health, these factors do not determine health and wellness. Sometimes health professionals construct genograms to document this information as illustrated in Figure 2.9 . There are various ways to construct genograms, this is one example.
You want to ask questions that elicit information about the health status, age, and if applicable, cause of death and age at time of death of blood relatives (parents, grandparents, siblings, children, nieces, nephews). Questions to ask include:
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Tell me about the health of your blood relatives.

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Have any of your blood relatives died?
- If so, do you know the cause of death?
- And at what age did they die?
You also want to assess the health status of non-blood relatives and individuals that clients have had close encounters with or live with; this can be an important part of the assessment particularly in terms of communicable diseases (i.e., diseases spread through air, food, physical contact, contaminated surfaces, and insect and animal bites). Questions to ask include:
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Tell me about the health status of those you live with.
- Has anyone been sick recently?
- If so, do you know the cause?
- What symptoms have they had?

Many people seek genetic information to better understand their family health. This information may be available from genetic counsellors, who are health professionals who offer informational and supportive resources for persons seeking to better understand their genetic health history and risk for genetic disorders and inherited conditions. Also popular today are the DNA tests available commercially to determine family ancestry and health predispositions. Despite their popularity, the accuracy and privacy risks of this latter type of testing is questionable.
Test Yourself
Query \(\PageIndex{1}\)